We know that $\frac{10}{1-10x}=10+100x+1000{{x}^{2}}+...+{{\left( 10 \right)}^{n+1}}{{x}^{n}}+...$ for $x\in \left(-\dfrac{1}{10},\dfrac{1}{10}\right)$. Using this fact, find the power series for $\dfrac{100}{(1-10x)^2}$. Choose 1 answer: Choose 1 answer: (Choice A) A $10+200x+3000{{x}^{2}}+...$ (Choice B) B $10-100x+1000{{x}^{2}}+...$ (Choice C) C $10+100x+1000{{x}^{2}}+...$ (Choice D) D $100+2000x+30000{{x}^{2}}+...$ (Choice E) E $-100+2000x-30000{{x}^{2}}+...$
Explanation: If we take the derivative with respect to $~x~$ of both sides of the given information, we obtain the result. ${{D}_{x}}\left( \frac{10}{1-10x} \right)={{D}_{x}}\left( 10+100x+1000{{x}^{2}}+...+{{\left( 10 \right)}^{n+1}}{{x}^{n}} \right)$ This gives the following equality. $\frac{100}{{{\left( 1-10x \right)}^{2}}}=100+2000x+30000{{x}^{2}}+...$